Let f(x)=max{6x,2+3x2}+∣x−1∣cosx2−41,x∈(−π,π).
The number of points at which f(x) is not differentiable is \_\_\_\_\_\_\_\_.
Answer: 3
Solutions · 1
Official solution
Part 1 --- max{6x,2+3x2}: The two expressions are equal when
\[2+3x^2=6x \implies 3x^2-6x+2=0 \implies x=\frac{6\pm\sqrt{12}}{6}=\frac{3\pm\sqrt{2}}{3}\]
At these two points the "active" branch of the max switches, and since dxd(6x)=6 while dxd(2+3x2)=6x are unequal there, the max function is not differentiable at both points.
Part 2 --- ∣x−1∣cosx2−41: At x=1, cos1−41=cos43=0. Since this cosine factor is nonzero at x=1, multiplying it by ∣x−1∣ does not smooth out the corner of ∣x−1∣, so this term is not differentiable at x=1.
Total number of points of non-differentiability =2+1=3.